Calculate the emf for the following reaction. Will the reaction occur spontaneously at 25°C, given that [Fe2+] = 0.600 M and [Cd2+] = 0.00550 M? Cd(s) + Fe2+(aq)→Cd2+(aq) + Fe(s) E o Cd2+/Cd = −0.40 V E o Fe2+/Fe = −0.44 V

The EMF of the cell is -0.100 V and the reaction is non-spontaneous

Explanation:

The half reaction for the equation follows:

Oxidation half reaction:

Reduction half reaction:

Net cell reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

where,

= electrode potential of the cell =  ?

= standard electrode potential of the cell = -0.04 V

n = number of electrons exchanged = 2

Putting values in above equation, we get:

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

Relationship between standard Gibbs free energy and standard electrode potential follows:

For a reaction to be spontaneous, the standard electrode potential must be positive.

Hence, the EMF of the cell is -0.100 V and the reaction is non-spontaneous

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i